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Quadrature rules with multiple nodes Gradimir V. Milovanović and Marija P. Stanić Abstract In this paper a brief historical survey of the development of quadrature rules with multiple nodes and the maximal algebraic degree of exactness is given. The natural generalization of such rules are quadrature rules with multiple nodes and the maximal degree of exactness in some functional spaces that are different from the space of algebraic polynomial. For that purpose we present a generalized quadrature rules considered by A. Ghizzeti and A. Ossicini [Quadrature Formulae, Academie–Verlag, Berlin, 1970] and apply their ideas in order to obtain quadrature rules with multiple nodes and the maximal trigonometric degree of exactness. Such quadrature rules are characterized by so called s− and σ −orthogonal trigonometric polynomials. Numerical method for the construction of such quadrature rules are given, as well as numerical the example which illustrate obtained theoretical results. 1 Introduction and preliminaries The most significant discovery in the field of numerical integration in 19th century was made by Carl Friedrich Gauß in 1814. His method [14] dramatically improved the earlier Newton’s method. Gauss’ method was enriched by significant contributions of Jacobi [21] and Christoffel [7]. More than hundred years after Gauß published his famous method there appeared the idea of numerical integration involving multiple nodes. Taking any system of n distinct points {τ1 , . . . , τn } and n nonnegative integers m1 , . . . , mn , and starting from the Hermite interpolation formula, L. Chakalov [4] in 1948 obtained the quadrature formula Gradimir V. Milovanović The Serbian Academy of Sciences and Arts, Kneza Mihaila 35, 11000 Belgrade, Serbia e-mail: [email protected] Marija P. Stanić Department of Mathematics and Informatics, Faculty of Science, University of Kragujevac, Radoja Domanovića 12, 34000 Kragujevac, Serbia e-mail: [email protected] 1 2 Gradimir V. Milovanović and Marija P. Stanić n Z 1 f (t) dt = −1 ∑ A0,ν f (τν ) + A1,ν f 0 (τν ) + · · · + Amν −1,ν f (mν −1) (τν ) , (1) ν=1 which is exact for all polynomials of degree at most m1 + · · · + mn − 1. He gave a method for computing the coefficients Ai,ν in (1). Such coefficients are given by R1 Ai,ν = −1 `i,ν (t) dt, ν = 1, . . . , n, i = 0, 1, . . . , mν − 1, where `i,ν (t) are the fundamental functions of Hermite interpolation. Taking m1 = · · · = mn = k in (1), P. Turán [48] in 1950 studied numerical quadratures of the form k−1 n Z 1 f (t) dt = −1 ∑ ∑ Ai,ν f (i) (τν ) + Rn,k ( f ), (2) i=0 ν=1 Obviously, for any given nodes −1 ≤ τ1 ≤ · · · ≤ τn ≤ 1 the formula (2) can be made exact for f ∈ Pkn−1 (Pm , m ∈ N0 , is the set of all algebraic polynomials of degree at most m). However, for k = 1 the formula (2), i.e., n Z 1 f (t) dt = −1 ∑ A0,ν f (τν ) + Rn,1 ( f ) ν=1 is exact for all polynomials of degree at most 2n − 1 if the nodes τν are the zeros of the Legendre polynomial Pn (it is the well–known Gauss–Legendre quadrature rule). Because of that it is quite natural to consider whether nodes τν can be chosen in such a way that the quadrature formula (2) will be exact for all algebraic polynomials of degree less that or equal to (k + 1)n − 1. Turán [48] showed that the answer is negative for k = 2, while it is positive for k = 3. He proved that the nodes τν , ν = 1, 2, . . . , n, should be chosen as the zeros of the monic polynomial πn∗ (t) = t n + · · · R1 which minimizes the integral −1 [πn (t)]4 dt, where πn (t) = t n + an−1t n−1 + · · · + a1t + a0 . In the general case, the answer is negative for even, and positive for odd k, when τν , ν = 1, 2, . . . , n, must be the zeros of the monic polynomial πn∗ minimizing R1 k+1 dt. Specially, for k = 1, π ∗ is the monic Legendre polynomial P bn . n −1 [πn (t)] Let us now assume that k = 2s + 1, s ≥ 0. Instead of (2), it is also interesting to consider a more general Gauss–Turán type quadrature formula 2s Z R f (t) dλ (t) = ∑ n ∑ Ai,ν f (i) (τν ) + Rn,2s ( f ), (3) i=0 ν=1 where dλ (t) is a given nonnegative measure on Rthe real line R, with compact or unbounded support, for which all moments µk = R t k dλ (t), k = 0, 1, . . ., exist and are finite, and µ0 > 0. It is known that formula (3) is exact for all polynomials of degree at most 2(s + 1)n − 1, i.e., Rn,2s ( f ) = 0 for f ∈ P2(s+1)n−1 . The nodes τν , ν = 1, . . . , n, in (3) are the zeros of the monic polynomial πn,s (t), which minimizes the integral Z [πn (t)]2s+2 dλ (t), F(a0 , a1 , . . . , an−1 ) = R Quadrature rules with multiple nodes 3 where πn (t) = t n + an−1t n−1 + · · · + a1t + a0 . This minimization leads to the conditions Z 1 ∂F · = [πn (t)]2s+1t k dλ (t) = 0, k = 0, 1, . . . , n − 1. (4) 2s + 2 ∂ ak R These polynomials πn = πn,s are known as s−orthogonal (or s−self associated) polynomials on R with respect to the measure dλ (t) (for more details see [15, 35, 36, 37]. For s = 0 they reduce to the standard orthogonal polynomials and (3) becomes the well–known Gauss–Christoffel formula. Using some facts about monosplines, Micchelli [22] investigated the sign of the Cotes coefficients Ai,ν in the Turán quadrature formula. For the numerical methods for the construction of s−orthogonal polynomials as well as for the construction of Gauss–Turán-type quadrature formulae we refer readers to the articles [23, 24, 17]. A next natural generalization of the Turán quadrature formula (3) is generalization to rules having nodes with different multiplicities. Such rule for dλ (t) = dt on (a, b) was derived independently by L. Chakalov [5, 6] and T. Popoviciu [38]. Important theoretical progress on this subject was made by D.D. Stancu [43, 44] (see also [46]). Let us assume that the nodes τ1 < τ2 < · · · < τn , have multiplicities m1 , m2 , . . . , mn , respectively (a permutation of the multiplicities m1 , m2 , . . . , mn , with the nodes held fixed, in general yields to a new quadrature rule). It can be shown that the quadrature formula (1) is exact for all polynomials of degree less than 2 ∑nν=1 [(mν + 1)/2]. Thus, the multiplicities mν that are even do not contribute toward an increase in the degree of exactness, so that it is reasonable to assume that all mν are odd integers, i.e., that mν = 2sν +1, ν = 1, 2, . . . , n. Therefore, for a given sequence of nonnegative integers σ = (s1 , s2 , . . . , sn ) the corresponding quadrature formula n 2sν Z f (t) dλ (t) = R ∑ ∑ Ai,ν f (i) (τν ) + R( f ) ν=1 i=0 has the maximal degree of exactness dmax = 2 ∑nν=1 sν + 2n − 1 if and only if Z n ∏ (t − τν )2sν +1t k dλ (t) = 0, k = 0, 1, . . . , n − 1. (5) R ν=1 The previous orthogonality conditions correspond to (4) and they could be obtained by the minimization of the integral Z n ∏ (t − τν )2sν +2 dλ (t). R ν=1 The existence of such quadrature rules was proved by Chakalov [5], Popoviciu [38], Morelli and Verna [33], and existence and uniqueness by Ghizzetti and Ossicini [18]. 4 Gradimir V. Milovanović and Marija P. Stanić The conditions (5) define a sequence of polynomials {πn,σ }n∈N0 , n (n,σ ) ∏ (t − τν πn,σ (t) = (n,σ ) ), τ1 (n,σ ) < τ2 (n,σ ) < · · · < τn , ν=1 such that n Z R (n,σ ) 2sν +1 πk,σ (t) ∏ (t − τν ) dλ (t) = 0, k = 0, 1, . . . , n − 1. ν=1 These polynomials πk,σ are called σ -orthogonal polynomials, and they correspond to the sequence σ = (s1 , s2 , . . .). Specially, for s = s1 = s2 = · · · , the σ −orthogonal polynomials reduce to the s−orthogonal polynomials. At the end of this section we mention a general problem investigated by Stancu [43, 44, 46]. Namely, let η1 , . . . , ηm (η1 < · · · < ηm ) be given fixed (or prescribed) nodes, with multiplicities `1 , . . . , `m , respectively, and τ1 , . . . , τn (τ1 < · · · < τn ) be free nodes, with given multiplicities m1 , . . . , mn , respectively. Stancu considered interpolatory quadrature formulae of a general form Z I( f ) = f (t) dλ (t) ∼ = R n mν −1 ∑ ∑ Ai,ν f (i) (τν ) + ν=1 i=0 m `ν −1 ∑ ∑ Bi,ν f (i) (ην ), (6) ν=1 i=0 with an algebraic degree of exactness at least M + L − 1, where M = ∑nν=1 mν and L = ∑m ν=1 `ν . Using free and fixed nodes we can introduce two polynomials n QM (t) : = ∏ (t − τν )mν m and ν=1 qL (t) : = ∏ (t − ην )`ν . ν=1 Choosing the free nodes to increase the degree of exactness leads to the so–called Gaussian type of quadratures. If the free (or Gaussian) nodes τ1 , . . . , τn are such that the quadrature rule (6) is exact for each f ∈ PM+L+n−1 , then we call it the Gauss– Stancu formula (see [26]). Stancu [45] proved that τ1 , . . . , τn are the Gaussian nodes if and only if Z t k QM (t)qL (t) dλ (t) = 0, k = 0, 1, . . . , n − 1. (7) R Under some restrictions of node polynomials qL (t) and QM (t) on the support interval of the measure dλ (t) we can give sufficient conditions for the existence of Gaussian nodes (cf. Stancu [45] and [20]). For example, if the multiplicities of the Gaussian nodes are odd, e.g., mν = 2sν + 1, ν = 1, . . . , n, and if the polynomial with fixed nodes qL (t) does not change its sign in the support interval of the measure dλ (t), then, in this interval, there exist real distinct nodes τν , ν = 1, . . . , n. This condition for the polynomial qL (t) means that the multiplicities of the internal fixed nodes must be even. Defining a new (nonnegative) measure dλ̂ (t) := |qL (t)| dλ (t), Quadrature rules with multiple nodes 5 the “orthogonality conditions” (7) can be expressed in a simpler form Z t k QM (t) dλ̂ (t) = 0, k = 0, 1, . . . , n − 1. R This means that the general quadrature problem (6), under these conditions, can be reduced to a problem with only Gaussian nodes, but with respect to another modified measure. Computational methods for this purpose are based on Christoffel’s theorem and described in details in [16] (see also [20]). For more details about the concept of power orthogonality and the corresponding quadrature with multiple nodes and the maximal algebraic degree of exactness and about the numerical methods for their construction we refer readers to the article [25] and the references therein, as well as to a nice book by Shi [41]. The next natural generalization represents quadrature rules with multiple nodes and the maximal degree of exactness in some functional spaces which are different from the space of algebraic polynomials. 2 A generalized Gaussian problem For a finite interval [a, b] and a positive integer N, Ghizzeti and Osicini in [19] conR sidered quadrature rules for the computation of an integral of the type ab f (x) w(x) dx, where the weight function w and function f satisfy the following conditions: w(x) ∈ L[a, b], f (x) ∈ ACN−1 [a, b] (ACm [a, b], m ∈ N0 , is the class of functions f whose m-th derivative is absolutely continuous in [a, b]). For a fixed number m ≥ 1 of points a ≤ x1 < x2 < · · · < xm ≤ b and a fixed linear differential operator E of order N of the form N E= dN−k ∑ ak (x) dxN−k , k=0 where a1 (x) = 1, ak (x) ∈ ACN−k−1 [a, b], k = 1, 2, . . . , N − 1, and aN (x) ∈ L(a, b), they considered (see [19, pp. 41–43]) a generalized Gaussian problem, i.e., they studied the quadrature rule m N−1 Z b f (x) w(x) dx = a ∑ ∑ A j,ν , f ( j) (xν ) + R( f ) ν=1 j=0 such that E( f ) = 0 implies R( f ) = 0. According to [19, p. 28] quadrature rule which is exact for all algebraic polynomials of degree less than or equal ν − 1 are relative to the differential operator dν E = dx ν , and quadrature rule which is exact for all trigonometric polynomials of degree less than or equal to ν are relative to the differential operator (of order 2ν + 1): d d2 2 E = dx ∏νk=1 dx 2 +k . 6 Gradimir V. Milovanović and Marija P. Stanić The considered whether there can exist a rule of the form m N−pν −1 Z b f (x)w(x) dx = a ∑ ∑ ν=1 A j,ν f ( j) (xν ) + R( f ), (8) j=0 with fixed integers pν , 0 ≤ pν ≤ N − 1, ν = 1, 2, . . . , m, such that at least one of the integers pν is greater than or equal to 1, satisfying that E( f ) = 0 implies R( f ) = 0, too. The answer is given in the following theorem (see [19, p. 45] for the proof). Theorem 1. For the given nodes x1 , x2 , . . . , xm , the linear differential operator E of order N and the nonnegative integers p1 , p2 , . . . , pm , 0 ≤ pν ≤ N −1, ν = 1, 2, . . . , m, such that there exists ν ∈ {1, 2, . . . , m} for which pν ≥ 1, consider the following homogenous boundary differential problem E( f ) = 0, f ( j) (xν ) = 0, j = 0, 1, . . . , N − pν − 1, ν = 1, 2, . . . , m. (9) If this problem has no non–trivial solutions (whence N ≤ mN − ∑m ν=1 pν ) it is posp − N parameters sible to write a quadrature rule of the type (8) with mN − ∑m ν ν=1 chosen arbitrarily. If, on the other hand, the problem (9) has q linearly independent solutions Ur , r = 1, 2, . . . , q, with N −mN + ∑m ν=1 pν ≤ q ≤ pν for all ν = 1, 2, . . . , m, then the formula (8) may apply only if the following q conditions Z b a Ur (x) w(x) dx = 0, r = 1, 2, . . . , q, are satisfied; if so, mN − ∑m ν=1 pν − N + q parameters in the formula (8) can be chosen arbitrary. Since the quadrature rules with multiple nodes and the maximal algebraic degree of exactness are widespread in literature, in what follows we restrict our attention to the quadrature rules with multiple nodes and the maximal trigonometric degree of exactness. 3 Quadrature rules with multiple nodes and the maximal trigonometric degree of exactness For a nonnegative integer n and for γ ∈ {0, 1/2} by Tn we denote the linear span of the set {cos(k + γ)x, sin(k + γ)x : k = 0, 1, . . . , n}. It is obvious that Tn0 = Tn is the linear space of all trigonometric polynomials of degree less than or equal to n, 1/2 Tn is the linear space of all trigonometric polynomials of semi–integer degree less S γ γ than or equal to n + 1/2 and dim(Tn ) = 2(n + γ) + 1. By T γ = n∈N0 Tn we denote the set of all trigonometric polynomials (for γ = 0) and the set of all trigonometric polynomials of semi–integer degree (for γ = 1/2). For γ = 0 we simple write T en we denote the linear space Tn span{sin nx} or Tn instead of T 0 . Finally, by T span{cos nx}. γ Quadrature rules with multiple nodes 7 We use the following notation: γb = 1 − 2γ, γ ∈ {0, 1/2}. In what follows we simple say that trigonometric polynomial has degree k + γ, k ∈ N0 , γ ∈ {0, 1/2}, which always means that for γ = 0 it is trigonometric polynomial of precise degree k, while for γ = 1/2 it is trigonometric polynomial of precise semi–integer degree k + 1/2. Obviously, every trigonometric polynomial of degree n + γ can be represented in the form n Tnγ (x) = ∑ (ck cos(k + γ)x + dk sin(k + γ)x) , c2n + dn2 6= 0, (10) k=0 where ck , dk ∈ R, k = 0, 1, . . . , n. For γ = 0 we always set d0 = 0. Coefficients cn and dn are called the leading coefficients. Let us suppose that w is a weight function, integrable and nonnegative on the interval [−π, π), vanishing there only on a set of a measure zero. The quadrature rules with simple nodes and the maximal trigonometric degree of exactness were considered by many authors. For a brief historical survey of available approaches for the construction of such quadrature rules we refer readers to the article [28] and the references therein. Following Turetzkii’s approach (see [49]), which is a simulation of the development of Gaussian quadrature rules for algebraic polynomials, quadrature rules with the maximal trigonometric degree of exactness with an odd number of nodes (γ = 1/2) and with an even number of nodes (γ = 0) were considered in [28] (see also [9]) and in [47], respectively. It is well known that in a case of quadrature rule with maximal algebraic degree of exactness the nodes are the zeros of the corresponding orthogonal algebraic polynomial. In a case of quadrature with an odd maximal trigonometric degree of exactness the nodes are the zeros of the corresponding orthogonal trigonometric polynomials, but for an even maximal trigonometric degree of exactness the nodes are not zeros of trigonometric polynomial, but zeros of the corresponding orthogonal trigonometric polynomial of semi–integer degree. Similarly, when the quadrature rules with multiple nodes and the maximal algebraic degree of exactness are in question it is necessary to consider power orthogonality in the space of algebraic polynomials. But, if we want to consider quadrature rules with multiple nodes and the maximal trigonometric degree of exactness, we have to consider power orthogonality in the space of trigonometric polynomials or in the space of trigonometric polynomials of semi–integer degree, which depends on number of nodes. In mentioned papers [28] and [47] the existence and the uniqueness of orthogonal trigonometric polynomials of integer or of semi–integer degree were proved under the assumption that two of the coefficients in expansion (10) are given in advance (it is usual to choose the leading coefficients in advance). Also, if we directly compute the zeros of orthogonal trigonometric polynomial (of integer or semi–integer degree), we can fix one of them in advance since for 2(n + γ) zeros we have 2(n + γ) − 1 orthogonality conditions. Let n be positive integer, γ ∈ {0, 1/2}, sν nonnegative integers for ν = γb, γb + 1, . . . , 2n, and σ = (sγb, sγb+1 , . . . , s2n ). For the given weight function w we considered 8 Gradimir V. Milovanović and Marija P. Stanić a quadrature rule of the following type 2n 2sν Z π f (x) w(x) dx = −π ∑ ∑ A j,ν f ( j) (xν ) + Rn ( f ). (11) ν=b γ j=0 which have maximal trigonometric degree of exactness, i.e., for which Rn ( f ) = 0 for all f ∈ TN1 , where N1 = ∑2n ν=b γ (sν + 1) − 1. In this case boundary differential problem (9) has the following form E( f ) = 0, f ( j) (xν ) = 0, j = 0, 1, . . . , 2sν , ν = γb, γb + 1, . . . , 2n, (12) where E is the following differential operator of order N = 2N1 + 1: d N1 ∏ dx k=1 E= d2 2 + k . dx2 The boundary problem (12) has the following 2n − γb linear independent nontrivial solutions 2n x − xν 2sν +1 U` (x) = ∏ sin cos(` + γ)x, ` = 0, 1, . . . , n − 1, 2 ν=b γ 2n x − xν 2sν +1 V` (x) = ∏ sin sin(` + γ)x, ` = γb, γb + 1, . . . , n − 1. 2 ν=b γ According to Theorem 1, the nodes xγb, xγb+1 , . . . , x2n of the quadrature rule (11) satisfy conditions x − xν 2sν +1 cos(` + γ)x w(x) dx = 0, ` = 0, 1, . . . , n − 1, ∏ sin 2 −π ν=b γ Z π 2n x − xν 2sν +1 sin sin(` + γ)x w(x) dx = 0, ` = γb, γb + 1, . . . , n − 1, ∏ 2 −π ν=b γ Z π 2n i.e., Z π 2n ∏ −π ν=b γ sin x − xν 2 2sν +1 γ tn−1 (x) w(x) dx = 0, for all tn−1 ∈ Tn−1 . γ γ (13) Trigonometric polynomial Tσ ,n = ∏2n ν=b γ sin(x − xν )/2 ∈ Tn which satisfies orthogonality conditions (13) is called σ −orthogonal trigonometric polynomial of degree n + γ with respect to weight function w on [−π, π). Therefore, the nodes of quadrature rule (11) with the maximal trigonometric degree of exactness are the zeros of the corresponding σ −orthogonal trigonometric polynomial. γ γ Quadrature rules with multiple nodes to 9 Specially, for sγb = sγb+1 = · · · = s2n = s, σ −orthogonality conditions (13) reduce Z π −π 2n x − xν ∏ sin 2 ν=b γ !2s+1 tn−1 ∈ Tn−1 . γ γ tn−1 (x) w(x) dx = 0, γ (14) Trigonometric polynomial Ts,n = ∏2n ν=b γ sin(x − xν )/2 ∈ Tn which satisfies (14) is called s−orthogonal trigonometric polynomial of degree n + γ with respect to w on [−π, π). For γ = 1/2, s−orthogonal trigonometric polynomials of semi–integer degree were defined and considered in [27], while σ −orthogonal trigonometric polynomials of semi–integer degree were defined and their main properties were proved in [29]. For γ = 0, the both s− and σ −orthogonal trigonometric polynomials were defined and studied in [47]. It is obvious that s−orthogonal trigonometric polynomials can be considered as a special case of σ −orthogonal trigonometric polynomials. However, we first prove the existence and uniqueness of s−orthogonal trigonometric polynomials because it can be done in a quite simple way, by using the well–known facts about the best approximation given in the Remark 1 below (see [10, p. 58–60]). The proofs of the existence and uniqueness of σ −orthogonal trigonometric polynomials are more complicated. We prove them by using theory of implicitly defined orthogonality (see [13, Sect. 5.3] for implicitly defined orthogonal algebraic polynomials). γ γ Remark 1. Let X be a Banach space and Y be a closed linear subspace of X. For each f ∈ X, infg∈Y k f − gk is the error of approximation of f by elements from Y . If there exists some g = g0 ∈ Y for which that infimum is attained, then g0 is called the best approximation to f from Y . For each finite dimensional subspace Xn of X and each f ∈ X, there exists the best approximation to f from Xn . In addition, if X is a strictly convex space, then each f ∈ X has at most one element of the best approximation in each closed linear subspace Y ⊂ X. Analogously as it was proved in [28] and [47] for orthogonal trigonometric polynomials, one can prove the existence and uniqueness of s− and σ −orthogonal trigonometric polynomials fixing in advance two of their coefficients or one of their zeros. Before we prove the existence and uniqueness of s− and σ −orthogonal trigonometric polynomials, we prove that all of their zeros in [−π, π) are simple. Of course, it is enough to prove that for σ −orthogonal trigonometric polynomials. Theorem 2. The σ −orthogonal trigonometric polynomial Tσ ,n ∈ Tn with respect to the weight function w(x) on [−π, π) has in [−π, π) exactly 2(n + γ) distinct simple zeros. γ γ γ Proof. It is easy to see that the trigonometric polynomial Tσ ,n (x) must have at least one zero of odd multiplicity in [−π, π), because if we assume the contrary, then for n ∈ N we obtain that Z π 2n ∏ −π ν=b γ sin x − xν 2sν +1 cos(0 + γ)x w(x) dx = 0, 2 10 Gradimir V. Milovanović and Marija P. Stanić which is impossible, because the integrand does not change its sign on [−π, π). We γ know (see [1] and [2]) that Tσ ,n (x) must change its sign an even number of times for γ = 0 and an odd number of times for γ = 1/2. First, we suppose that γ = 0 and that the number of zeros of odd multiplicities of Tσ ,n (x) on [−π, π) is 2m, m < n. We denote these zeros by y1 , y2 , . . . , y2m , and set t(x) = ∏2m k=1 sin((x − yk )/2). Since t ∈ Tm , m < n, we have Z π 2n sin ∏ −π ν=1 x − xν 2sν +1 t(x) w(x) dx = 0, 2 which also gives a contradiction, since the integrand does not change its sign on [−π, π). Thus, Tσ ,n must have exactly 2n distinct simple zeros on [−π, π). Similarly, for γ = 1/2 by y1 , y2 , . . . , y2m+1 , m < n, we denote the zeros of odd 1/2 multiplicities of Tσ ,n (x) on [−π, π) and set b t (x) = ∏2m+1 k=1 sin((x − yk )/2). Since 1/2 b t ∈ Tm , m < n, we have Z π 2n ∏ −π ν=0 sin x − xν 2sν +1 (x)b t (x) w(x) dx = 0, 2 which again gives a contradiction, since the integrand does not change its sign on 1/2 [−π, π), which means that Tσ ,n (x) must have exactly 2n + 1 distinct simple zeros on [−π, π). t u 3.1 s−orthogonal trigonometric polynomials Theorem 3. Trigonometric polynomial Ts,n (x) ∈ Tn , γ ∈ {0, 1/2}, with given leading coefficients, which is s–orthogonal on [−π, π) with respect to a given weight function w is determined uniquely. γ γ Proof. Let us set X = L2s+2 [−π, π], u = w(x)1/(2s+2) (cn,γ cos(n + γ)x + dn,γ sin(n + γ)x) ∈ L2s+2 [−π, π], and fix the following 2n − γb linearly independent elements in L2s+2 [−π, π]: u j = w(x)1/(2s+2) cos( j + γ)x, vk = w(x)1/(2s+2) sin(k + γ)x, j = 0, 1, . . . , n − 1, k = γb, γb + 1, . . . , n − 1. Then Y = span{u j , vk : j = 0, 1, . . . , n − 1, k = γb, γb + 1, . . . , n − 1} is a finite dimensional subspace of X and, according to Remark 1, for each element from X there exists the best approximation from Y , i.e., there exist 2n − γb constants α j,γ , j = 0, 1, . . . , n − 1, and βk,γ , k = γb, γb + 1, . . . , n − 1, such that the error Quadrature rules with multiple nodes 11 n−1 Z n−1 u − ∑ αk,γ uk + ∑ βk,γ vk = k=0 −π k=b γ n−1 − π cn,γ cos(n + γ)x + dn,γ sin(n + γ)x 2s+2 n−1 ∑ αk,γ cos(k + γ)x + ∑ βk,γ sin(k + γ)x k=0 1/(2s+2) w(x) dx , k=b γ is minimal, i.e., for every n and for every choice of the leading coefficients cn,γ , dn,γ , 2 6= 0, there exists a trigonometric polynomial of degree n + γ c2n,γ + dn,γ γ Ts,n (x) = cn,γ cos(n + γ)x + dn,γ sin(n + γ)x n−1 n−1 − ∑ αk,γ cos(k + γ)x + ∑ βk,γ sin(k + γ)x , k=0 such that k=b γ Z π −π γ (Ts,n (x))2s+2 w(x) dx is minimal. Since the space L2s+2 [−π, π] is strictly convex, according to Remark 1, the problem of the best approximation has the unique solution, i.e., the trigonometric γ polynomial Ts,n is unique. Therefore, for each of the following 2n − γb functions FkC (λ ) = FkS (λ ) = Z π −π Z π −π γ Ts,n (x) + λ cos(k + γ)x γ Ts,n (x) + λ sin(k + γ)x 2s+2 2s+2 w(x) dx, k = 0, 1, . . . , n − 1, w(x) dx, k = γb, γb + 1, . . . , n − 1, its derivative must be equal to zero for λ = 0. Hence, Z π −π Z π −π 2s+1 γ (x) cos(k + γ)x w(x) dx = 0, Ts,n k = 0, 1, . . . , n − 1, 2s+1 γ Ts,n (x) sin(k + γ)x w(x) dx = 0, k = γb, γb + 1, . . . , n − 1, γ which means that the polynomial Ts,n (x) satisfies (14). t u 3.2 σ −orthogonal trigonometric polynomials In order to prove the existence and uniqueness of σ −orthogonal trigonometric polynomials we fix in advance the zero xγb = −π and consider the σ −orthogonality conditions (13) as the system with unknown zeros. Theorem 4. Let w be the weight function on [−π, π) and let p be a nonnegative continuous function, vanishing only on a set of a measure zero. Then there exists a 12 Gradimir V. Milovanović and Marija P. Stanić γ trigonometric polynomial Tn , γ ∈ {0, 1/2}, of degree n + γ, orthogonal on [−π, π) to every trigonometric polynomial of degree less than or equal to n − 1 + γ with γ respect to the weight function p Tn (x) w(x). bnγ we denote the set of all trigonometric polynomials of degree n + γ Proof. By T which have 2(n + γ) real distinct zeros −π = xγb < xγb+1 < · · · < x2n < π and denote S2n−bγ = x = (xγb+1 , xγb+2 , . . . , x2n ) ∈ R2n−bγ : −π < xγb+1 < xγb+2 < · · · < x2n < π . bn , we introduce the inner product as For a given function p and an arbitrary Qn ∈ T follows Z π f (x)g(x) p(Qγn (x))w(x) dx. f , g Qγ = γ n γ −π bnγ and S It is obvious that there is one to one correspondence between the sets T 2n−b γ. Indeed, for every element x = (xγb+1 , xγb+2 , . . . , x2n ) ∈ S2n−bγ and for Qγn (x) = cos x 2n x − xν sin ∏ 2 ν=bγ +1 2 (15) γ bγ , there exists a unique system of orthogonal trigonometric polynomials Uk ∈ T k γ γ k = 0, 1, . . . , n, such that Un , cos(k + γ)x Qγ = Un , sin( j + γ)x Qγ = 0, for all n n γ γ k = 0, 1, . . . , n − 1, j = γb, γb + 1, . . . , n − 1, and Un ,Un Qγ 6= 0. n In such a way, we introduce a mapping Fn : S2n−bγ → S2n−bγ , defined in the following way: for any x = (xγb+1 , xγb+2 , . . . , x2n ) ∈ S2n−bγ we have Fn (x) = y, where y = (yγb+1 , yγb+2 , . . . , y2n ) ∈ S2n−bγ is such that −π, yγb+1 , yγb+2 , . . . , y2n are the zeros of the orthogonal trigonometric polynomial of degree n + γ with respect to the weight γ γ function p(Qn (x))w(x), where Qn (x) is given by (15). It is easy to see that the func 2n tion p cos(x/2) ∏ν=bγ +1 sin((x − xν )/2) w(x) is also an admissible weight function for arbitrary x = (xγb+1 , xγb+2 , . . . , x2n ) ∈ S2n−bγ \S2n−bγ . Now, we prove that Fn is continuous mapping on S2n−bγ . Let x ∈ S2n−bγ be an arbitrary point, {x(m) }, m ∈ N, a convergent sequence of points from S2n−bγ , which converges to x, y = Fn (x), and y(m) = Fn (x(m) ), m ∈ N. Let y∗ ∈ S2n−bγ be an arbitrary limit point of the sequence {y(m) } when m → ∞. Thus, Quadrature rules with multiple nodes 13 (m) Z π cos −π x − yν x 2n sin ∏ 2 ν=bγ +1 2 cos(k + γ)x (m) x 2n x−x × p cos ∏ sin 2 ν 2 ν=b γ +1 ! w(x) dx = 0, k = 0, 1, . . . , n − 1, w(x) dx = 0, j = γb, γb + 1, 2, . . . , n − 1. (m) Z π cos −π x−y x 2n ∏ sin 2 ν 2 ν=b γ +1 sin( j + γ)x (m) x 2n x−x × p cos ∏ sin 2 ν 2 ν=b γ +1 ! Applying Lebesgue Theorem of dominant convergence (see [32, p. 83]), when m → +∞ we obtain Z π cos −π Z π cos −π x − y∗ν x 2n sin cos(k + γ)x ∏ 2 ν=bγ +1 2 ! x 2n x − xν p cos ∏ sin 2 w(x) dx = 0, 2 ν=b γ +1 x 2n x − y∗ν sin sin( j + γ)x ∏ 2 ν=bγ +1 2 ! x − xν x 2n × p cos ∏ sin 2 w(x) dx = 0, 2 ν=b γ +1 k = 0, 1, . . . , n − 1, j = γb, γb + 1, . . . , n − 1, ∗ i.e., cos(x/2) ∏2n ν=b γ +1 sin((x − yν )/2) is the trigonometric polynomial of degree n + γ which is orthogonal to all trigonometric polynomials from Tn−1 with respect to the weight function p cos(x/2) ∏2n ν=b γ +1 sin((x − xν )/2) w(x) on [−π, π). Such trigonometric polynomial has 2(n + γ) distinct simple zeros in [−π, π) (see [28] and [47]). Therefore, y∗ ∈ S2n−bγ and y∗ = Fn (x). Since y = Fn (x), because of the uniqueness we have y∗ = y, i.e., the mapping Fn is continuous on S2n−bγ . Finally, we prove that the mapping Fn has a fixed point. The mapping Fn is continuous on the bounded, convex and closed set S2n−bγ ⊂ R2n−bγ . Applying the Brouwer fixed point theorem (see [34]) we conclude that there exists a fixed point of Fn . Since Fn (x) ∈ S2n−bγ for all x ∈ S2n−bγ \S2n−bγ , the fixed point of Fn belongs to S2n−bγ . If we denote the fixed point of Fn by x = (xγb+1 , xγb+2 , . . . , x2n ), then for Tn (x) = cos(x/2) ∏2n ν=b γ +1 sin((x − xν )/2) we get γ Z π −π Z π −π Tnγ (x) cos(k + γ)x p Tnγ (x) w(x) dx = 0, k = 0, 1, . . . , n − 1, Tnγ (x) sin( j + γ)x p Tnγ (x) w(x) dx = 0, j = γb, γb + 1, . . . , n − 1. t u 14 Gradimir V. Milovanović and Marija P. Stanić 3.3 Construction of quadrature rules with multiple nodes and the maximal trigonometric degree of exactness The construction of quadrature rule with multiple nodes and the maximal trigonometric degree of exactness is based on the application of some properties of the topological degree of mapping (see [39]). For that purposes we use a certain modification of ideas of Bojanov [3], Shi [40], Shi and Xu [42] (for algebraic σ −orthogonal polynomials) and Dryanov [11]. Using a unique notation for γ ∈ {0, 1/2} and omitting some details we present here results obtained in [29] and [47] for γ = 1/2 and γ = 0, respectively. For a ∈ [0, 1], n ∈ N, and F(x, a) = x 2asγb+1 2n x − xν 2asν +1 cos ∏ sin 2 2 ν=b γ +1 ! 2n x − xν γ × sgn ∏ sin tn−1 (x) w(x) dx, 2 ν=b γ +1 Z π −π tn−1 ∈ Tn−1 , γ γ we consider the following problem F(x, a) = 0 for all tn−1 ∈ Tn−1 , γ γ (16) with unknowns xγb+1 , xγb+2 , . . . , x2n . The σ –orthogonality conditions (13) (with x1 = −π) are equivalent to the problem (16) for a = 1, which means that the nodes of the quadrature rule (11) can be obtained as a solution of the problem (16) for a = 1. For a = 0 and a = 1 the problem (16) has the unique solution in the simplex S2n−bγ . According to Theorem 4 we know that the problem (16) has solutions in the simplex S2n−bγ for every a ∈ [0, 1]. We will prove the uniqueness of the solution x ∈ S2n−bγ of the problem (16) for all a ∈ (0, 1). Let us denote 2n x − xν 2asν +2 W (x, a, x) = ∏ sin 2 ν=b γ +1 and φk (x, a) = Z π cos −π x 2asγb+1 W (x, a, x) x − xk w(x) dx, 2 sin 2 (17) for k = γb + 1, γb + 2, . . . , 2n. Applying the same arguments as in [11, Lemma 3.2], the following auxiliary result can be easily proved. Lemma 1. There exists ε > 0 such that for every a ∈ [0, 1] the solutions x of the problem (16) belong to the simplex Sε,γ = y : ε ≤ yγb+1 + π, ε ≤ yγb+2 − yγb+1 , . . . , ε ≤ y2n − y2n−1 , ε ≤ π − y2n . Quadrature rules with multiple nodes 15 Also, the following result holds (see [29, Lemma 2.3] and [47, Lemma 4.4] for the proof). Lemma 2. The problem (16) and the following problem φk (x, a) = 0, k = γb + 1, γb + 2, . . . , 2n, (18) where φk (x, a) are given by (17), are equivalent in the simplex S2n−bγ . It is important to emphasize that the problems (16) and (18) are equivalent in the simplex Sε,γ , for some ε > 0, as well as in the simplex Sε1 ,γ , for all 0 < ε1 < ε, but they are not equivalent in S2n−bγ . For the proof of the main result we need the following lemma (for the proof see [29, Lemma 2.4]). Lemma 3. Let pξ ,η (x) be a continuous function on [−π, π], which depends continuously on parameters ξ , η ∈ [c, d], i.e., if (ξm , ηm ) approaches (ξ0 , η0 ) then the sequences pξm ,ηm (x) tends to pξ0 ,η0 (x) for every fixed x. If the solution x(ξ , η) of the problem (18) with the weight function pξ ,η (x)w(x) is always unique for every (ξ , η) ∈ [c, d]2 , then the solution x(ξ , η) depends continuously on (ξ , η) ∈ [c, d]2 . Now, we are ready to prove the main result. Theorem 5. The problem (18) has a unique solution in the simplex S2n−bγ for all a ∈ [0, 1]. Proof. We call the problem (18) as (a; sγb+1 , sγb+2 , . . . , s2n ; w) problem and prove this theorem by mathematical induction on n. The uniqueness for n = 0 is trivial. As an induction hypothesis, we suppose that the (a; sγb+1 , sγb+2 , . . . , s2n−2 ; w) problem has a unique solution for every a ∈ [0, 1] and for every weight function w, integrable and nonnegative on the interval [−π, π), vanishing there only on a set of a measure zero. For (ξ , η) ∈ [−π, π]2 we define the weight functions x − ξ 2as2n−1 +2 x − η 2as2n +2 w(x). pξ ,η (x) = sin sin 2 2 According to the induction hypothesis, the (a; sγb+1 , sγb+2 , . . . , s2n−2 ; pξ ,η ) problem has a unique solution (xγb+1 (ξ , η), xγb+2 (ξ , η), . . . , x2n−2 (ξ , η)) for every a ∈ [0, 1], such that −π < xγb+1 (ξ , η) < · · · < x2n−2 (ξ , η) < π, and xν (ξ , η) for all ν = γb + 1, γb + 2, . . . , 2n − 2 depends continuously on (ξ , η) ∈ [−π, π]2 (see Lemma 3). We will prove that the solution of the (a; sγb+1 , sγb+2 , . . . , s2n ; w) problem is unique for every a ∈ [0, 1]. Let us denote x2n−1 (ξ , η) = ξ , x2n (ξ , η) = η, x − xν (ξ , η) 2asν +2 W (x(ξ , η), a, x) = ∏ sin , 2 ν=b γ +1 2n 16 Gradimir V. Milovanović and Marija P. Stanić and φk (x(ξ , η), a) = Z π cos −π x 2asγb+1 W (x(ξ , η), a, x) w(x) dx, x − xk (ξ , η) 2 sin 2 k = γb + 1, . . . , 2n. The induction hypothesis gives us that φk (x(ξ , η), a) = 0, k = γb + 1, γb + 2, . . . , 2n − 2, (19) for (ξ , η) ∈ D, where D = {(ξ , η) : x2n−2 (ξ , η) < ξ < η < π}. Let us now consider the following problem in D with unknown t = (ξ , η): x 2asγb+1 W (x(ξ , η), a, x) w(x) dx = 0, x−ξ 2 −π sin 2 Z π x 2asγb+1 W (x(ξ , η), a, x) ϕ2 (t, a) = cos w(x) dx = 0. x−η 2 −π sin 2 ϕ1 (t, a) = Z π cos (20) If (ξ , η) ∈ D is a solution of the problem (20), then, according to (19), x(ξ , η) is a solution of the (a; sγb+1 , . . . , s2n ; w) problem in the simplex S2n−bγ . To the contrary, if x = (xγb+1 , xγb+2 , . . . , x2n ) is a solution of the (a; sγb+1 , . . . , s2n ; w) problem in the simplex S2n−bγ , then (x2n−1 , x2n ) is a solution of the problem (20). Applying Lemma 1 we conclude that every solution of the problem (20) belongs to Dε = {(ξ , η) : ε ≤ ξ − x2n−2 (ξ , η), ε ≤ η − ξ , ε ≤ π − η}, for some ε > 0. Let x be a solution of the problem (a; sγb+1 , sγb+2 , . . . , s2n ; w) in Sε,γ . Then, differentiating ϕ1 with respect to the xk , for all k = γb + 1, γb + 2, . . . , 2n − 2, 2n, we have x 2asγb+1 2n x − xν 2asν +1 cos ∏ sin 2 2 −π ν=b γ +1 ! 2n 2n x − xν x − xν x − xk × sgn ∏ sin sin cos w(x) dx. ∏ 2 2 2 ν=b γ +1 ν=b γ +1 ∂ ϕ1 = − (ask + 1) ∂ xk Z π ν6=k, ν6=2n−1 Since 2n ∏ ν=b γ +1 ν6=k, ν6=2n−1 sin x − xν x − xk γ cos ∈ Tn−1 , 2 2 applying Lemma 2, we conclude that ∂∂ϕx 1 = 0, for all k = γb + 1, . . . , 2n − 2, 2n. Furk ther, applying elementary trigonometric transformations we get the following identity x−y x y x−y y y x−y 2 y cos = cos cos + sin cos sin + cos sin , 2 2 2 2 2 2 2 2 and for Quadrature rules with multiple nodes 17 2as2n−1 + 1 ∂ ϕ1 =− I := ∂ x2n−1 2 x 2asγb+1 cos 2 2n Z π −π ∏ ν=b γ +1 ν6=2n−1 x − xν 2asν +2 sin 2 x − x2n−1 2as2n−1 x − x2n−1 cos × sin w(x) dx, 2 2 we obtain cos x2n−1 x2n−1 2as2n−1 + 1 I1 + sin I=− I2 , 2 2 2 (21) where I1 = I2 = x 2asγb+1 cos 2 2n Z π −π x 2asγb+1 cos 2 ∏ ν=b γ +1 ν6=2n−1 2n Z π −π ∏ ν=b γ +1 ν6=2n−1 x − xν 2asν +2 x − x2n−1 2as2n−1 x sin sin cos w(x) dx, 2 2 2 x − xν 2asν +2 sin 2 x − x2n−1 2as2n−1 x − x2n−1 × sin sin w(x) dx. 2 2 Obviously, I1 > 0 (the integrand does not change its sign on [−π, π)) and I2 = 0 (because of (20)). Since −π < x2n−1 < π, from (21) we get ∂ ϕ1 sgn = −1. ∂ x2n−1 Analogously, ∂ ϕ2 = 0, ∂ xk k = γb + 1, γb + 2, . . . , 2n − 1, ∂ ϕ2 sgn ∂ x2n = −1. Thus, at any solution t = (ξ , η) of the problem (20) from Dε we have ∂ ϕ1 = ∂ξ 2n ∑ k=b γ +1 k6=2n−1 ∂ ϕ1 ∂ ϕ1 ∂ ϕ1 ∂ xk · + = , ∂ xk ∂ ξ ∂ x2n−1 ∂ x2n−1 2n−1 ∂ ϕ2 ∂ xk ∂ ϕ2 ∂ ϕ2 ∂ ϕ2 = ∑ · + = , ∂ η k=bγ +1 ∂ xk ∂ η ∂ x2n ∂ x2n ∂ ϕ1 = 0, ∂η ∂ ϕ2 = 0. ∂ξ Hence, the determinant of the corresponding Jacobian J(t) is positive. We are going to prove that the problem (20) has a unique solution in Dε/2 . We prove it by using the topological degree of a mapping. Let us define the mapping ϕ(t, a) : Dε/2 × [0, 1] → R2 , by ϕ(t, a) = (ϕ1 (t, a), ϕ2 (t, a)), t ∈ R2 , a ∈ [0, 1]. If x(ξ , η) is a solution of the (a; sγb+1 , . . . , s2n ; w) problem in Sε,γ , then the solutions of the problem ϕ(t, a) = (0, 0) in Dε/2 belong to Dε . The problem ϕ(t, a) = (0, 0) has 18 Gradimir V. Milovanović and Marija P. Stanić the unique solution in Dε/2 for a = 0. It is obvious that ϕ( · , 0) is differentiable on Dε/2 , the mapping ϕ(t, a) is continuous in Dε/2 × [0, 1], and ϕ(t, a) 6= (0, 0) for all t ∈ ∂ Dε/2 and a ∈ [0, 1]. Then deg(ϕ( · , a), Dε/2 , (0, 0)) = sgn(det(J(t))) = 1, for all a ∈ [0, 1], which means that deg(ϕ( · , a), Dε/2 , (0, 0)) is a constant independent of a. Therefore, the problem ϕ(t, a) = (0, 0) has the unique solution in Dε/2 for all a ∈ [0, 1]. Hence, the (a; sγb+1 , . . . , s2n ; w) problem has a unique solution in S2n−bγ u which belongs to Sε,γ . t Theorem 6. The solution x = x(a) of the problem (18) depends continuously on a ∈ [0, 1]. Proof. Let {am }, am ∈ [0, 1], m ∈ N, be a convergent sequence, which converges to a∗ ∈ [0, 1]. Then for every am , m ∈ N, there exists the unique solution x(am ) = (xγb+1 (am ), . . . , x2n (am )) of the system (18) with a = am . The unique solution of sys∗ ) be an tem (18) for a = a∗ is x(a∗ ) = (xγb+1 (a∗ ), . . . , x2n (a∗ )). Let x∗ = (xγ∗b+1 , . . . , x2n arbitrary limit point of the sequence x(am ) when am → a∗ . According to Theorem 5, for each m ∈ N we have Z π x 2am sγb+1 W (x(am ), am , x) cos w(x) dx = 0, k = γb + 1, γb + 2, . . . , 2n. x − xk (am ) 2 −π sin 2 When am → a∗ the above equations lead to Z π cos −π x 2a∗ sγb+1 W (x∗ , a∗ , x) w(x) dx = 0, x − xk∗ 2 sin 2 k = γb + 1, γb + 2, . . . , 2n. According to Theorem 5 we get that x∗ = x(a∗ ), i.e., lim ∗ x(am ) = x(a∗ ). am →a t u 4 Numerical method for construction of quadrature rules with multiple nodes and the maximal trigonometric degree of exactness Based on the previously given theoretical results we present the numerical method for construction of quadrature rules of the form (11), for which Rn ( f ) = 0 for all f ∈ TN1 , where N1 = ∑2n ν=b γ (sν + 1) − 1. The first step is a construction of nodes, and the second one is a construction of weights knowing nodes. Quadrature rules with multiple nodes 19 4.1 Construction of nodes We chose xγb = −π and obtain the nodes xγb+1 , xγb+2 , · · · , x2n by solving system of nonlinear equations (18) for a = 1, applying Newton–Kantorovič method. Obtained theoretical results suggest that for fixed n the system (18) can be solved progressively, for an increasing sequence of values for a, up to the target value a = 1. The solution for some a(i) can be used as the initial iteration in Newton–Kantorovič method for calculating the solution for a(i+1) , a(i) < a(i+1) ≤ 1. If for some chosen a(i+1) Newton–Kantorovič method does not converge, we decrease a(i+1) such that it becomes convergent, which is always possible according to Theorem 6. Thus, we can set a(i+1) = 1 in each step, and, if that iterative process is not convergent, we set a(i+1) := (a(i+1) + a(i) )/2 until it becomes convergent. As the initial iteration for the first iterative process, for some a(1) > 0, we choose the zeros of the corresponding orthogonal trigonometric polynomial of degree n + γ, i.e., the solution of system (18) for a = 0. Let us introduce the following matrix notation T x = xγb+1 xγb+2 · · · x2n , h i (m) (m) (m) T x(m) = xγb+1 xγb+2 · · · x2n , m = 0, 1, . . . , T φ (x) = φγb+1 (x) φγb+2 (x) · · · φ2n (x) . Jacobian of φ (x), " ∂ φi+bγ W = W(x) = [wi, j ](2n−bγ )×(2n−bγ ) = ∂ x j+bγ # , (2n−b γ )×(2n−b γ) has the following entries ∂ φi = −(1 + as j ) ∂xj −π 2n × ∏ sin ν=b γ +1 ν6=i,ν6= j ∂ φi 1 + 2asi =− ∂ xi 2 x 2asγb+1 2n x − xν 2asν +1 sin cos sgn ∏ 2 2 ν=b γ +1 Z π x−xj x − xν cos w(x) dx, 2 2 2n x − xν ∏ sin 2 ν=b γ +1 ! i 6= j, i, j = γb + 1, γb + 2, . . . , 2n, x 2asγb+1 2n x − xν 2asν +2 cos ∏ sin 2 2 ν=b γ +1 Z π −π ν6=i x − xi 2asi cos x − xi w(x) dx, × sin 2 2 i = γb + 1, γb + 2, . . . , 2n. All of the above integrals can be computed by using a Gaussian type quadrature rule for trigonometric polynomials (see [28] and [47]) 20 Gradimir V. Milovanović and Marija P. Stanić 2N Z π f (x) w(x) dx = −π ∑ Aν f (xν ) + RN ( f ), ν=b γ with 2N ≥ ∑2n ν=b γ sν + 2n. The Newton–Kantorovič method for calculating the zeros of the σ –orthogonal γ trigonometric polynomial Tσ ,n is given as follows x(m+1) = x(m) − W−1 (x(m) )φ (x(m) ), m = 0, 1, . . . , and for sufficiently good chosen initial approximation x(0) , it has the quadratic convergence. 4.2 Construction of weights Knowing the nodes of quadrature rule (11), it is possible to calculate the corresponding weights. The weights can be calculating by using the Hermite trigonometric interpolation polynomial (see [11, 12]). Since the construction of the Hermite interpolation trigonometric polynomial is more difficult than in the algebraic case, we use an adaptation on the method which was first given in [17] for construction of Gauss–Turán quadrature rules (for algebraic polynomials) and then generalized for Chakalov–Popoviciu’s type quadrature rules (also for algebraic polynomials) in [30, 31]. For γ = 1/2 and γ = 0 the corresponding method for construction of weights of quadrature rule (11) is presented in [29] and [47], respectively. Our method is based on the facts that quadrature rule (11) is of interpolatory type and that it is exact for all trigonometric polynomials of degree less than or 2n equal to ∑2n ν=b γ (sν + 1) − 1 = ∑ν=b γ sν + 2(n + γ) − 1. Thus, the weights can be calculated requiring that quadrature rule (11) integrates exactly all trigonometric polynomials of degree less than or equal to ∑2n ν=b γ sν + (n + γ) − 1. Additionally, in the b case γ = 0 (i.e., γ = 1) we require that (11) integrates exactly cos ∑2n ν=1 sν + n x, 2n when ∑2n ν=1 (2sν + 1)xν = `π for an odd integer `, or sin ∑ν=1 sν + n x when 2n 2n ∑ν=1 (2sν + 1)xν = `π for an even integer `, while if ∑ν=1 ν 6= `π, `∈ Z, (2sν + 1)x 2n one can choose to require exactness for cos ∑2n s + n x or sin ∑ ν=1 ν ν=1 sν + n x arbitrary (see [1, 2, 47]). In such a way we obtain a system of linear equations for the unknown weights. That system can be solved by decomposing into a set of 2(n + γ) upper triangular systems. For γ = 0 we explain that decomposing method in the case when ∑2n ν=1 (2sν + 1)xν = `π, ` ∈ Z. Let us denote Quadrature rules with multiple nodes 21 x − xi 2si +1 Ων (x) = ∏ sin , ν = γb, γb + 1, . . . , 2n, 2 i=b γ 2n i6=ν x − xν k uk,ν (x) = sin Ων (x), 2 ν = γb, γb + 1, . . . , 2n, k = 0, 1, . . . , 2sν , and ( tk,ν (x) = uk,ν (x) cos uk,ν (x), x − xν , k−even, γ = 0, or k−odd, γ = 1/2, 2 k−odd, γ = 0, or k−even, γ = 1/2, (22) for all ν = γb, γb + 1, . . . , 2n, k = 0, 1, . . . , 2sν . It is easy to see that tk,ν is a trigonometric polynomial of degree less than or equal to ∑2n ν=b γ sν+ n, which for γ = 0 2n 2n has the leading term cos ∑ν=1 sν + n x or sin ∑ν=1 sν + n x, but not the both of them. Quadrature rule (11) is exact for all such trigonometric polynomials, i.e., Rn (tk,ν ) = 0 for all ν = γb, γb + 1, . . . , 2n, k = 0, 1, . . . , 2sν . ( j) By using the Leibniz differentiation formula it is easy to check that tk,ν (xi ) = 0, j = 0, 1, . . . , 2sν , for all i 6= ν. Hence, for all k = 0, 1, . . . , 2sν we have 2sν Z π µk,ν = −π tk,ν (x) w(x) dx = ( j) ∑ A j,ν tk,ν (xν ), j=0 ν = γb, γb + 1, . . . , 2n. (23) In such a way, we obtain 2(n + γ) independent systems for calculating the weights A j,ν , j = 0, 1, . . . , 2sν , ν = γb, γb+ 1, . . . , 2n. Here, we need to calculate the derivatives ( j) tk,ν (xν ), k = 0, 1, . . . , 2sν , j = 0, 1, . . . , 2sν , for each ν = γb, γb + 1, . . . , 2n. For that purpose we use the following result (see [29] for details). Lemma 4. For the trigonometric polynomials tk,ν , given by (22) we have (i) tk,ν (xν ) = 0, i < k; (k) tk,ν (xν ) = k! Ων (xν ), 2k and for i > k ( (i) tk,ν (xν ) = (i) vk,ν (xν ), k−even, γ = 0, or k−odd, γ = 1/2, (i) uk,ν (xν ), k−odd, γ = 0, or k−even, γ = 1/2, where (i) vk,ν (xν ) = [i/2] ∑ m=0 (i) i (−1)m (i−2m) u (xν ), 2m 22m k,ν and the sequence uk,ν (xν ), k ∈ N0 , i ∈ N0 , is the solution of the difference equation 22 (i) fk,ν Gradimir V. Milovanović and Marija P. Stanić [(i−1)/2] = i (−1)m (i−2m−1) f (xν ), ν = γb, γb+1, . . . , 2n, k ∈ N, i ∈ N0 , 2m + 1 22m+1 k−1,ν ∑ m=[(i−k)/2] (i) (i) with the initial conditions f0,ν = Ων (xν ). (i) Finally, the remaining problem of calculating Ων (xν ), ν = γb, γb + 1, . . . , 2n, is solved in [29, Lemma 3.3]. Lemma 5. Let us denote x − xi 2si +1 Ων,k (x) = ∏ sin . 2 i≤k i6=ν (i) Then Ων,ν−1 (x) = Ων,ν (x) and Ων (x) = Ων,2n (x). The sequence Ων,k,` (x), `, i ∈ N0 , where x − xk+1 ` Ων,k (x), k = γb, γb + 1, . . . , 2n − 1, k 6= ν − 1, Ων,k,` (x) = sin 2 is the solution of the following difference equation i (i) fν,k,` (x) = x − xk+1 mπ i 1 (i−m) sin + fν,k,`−1 (x), ∑ m 2m 2 2 m=0 (i) ` ∈ N0 , (i) with the initial conditions fν,k,0 = Ων,k , i ∈ N0 . The following equalities (i) (i) Ων,k,2sk+1 +1 (x) = Ων,k+1,0 (x), (i) (i) (i) (i) k 6= ν − 1, (i) Ων (x) = Ων,2n (x) = Ων,2n−1,2s2n +1 (x), ν 6= 2n, (i) Ω2n (x) = Ω2n,2n−1 (x) = Ω2n,2n−2,2s2n−1 +1 (x) hold. According to Lemma 4, the systems (23) are the following upper triangular systems 0 (x ) · · · t (2sν ) (x ) t0,ν (xν ) t0,ν ν ν A0,ν µ0,ν 0,ν µ 0 (x ) · · · t (2sν ) (x ) A t 1,ν 1,ν ν ν 1,ν 1,ν · = , ν = γb, γb + 1, . . . , 2n. . . . .. .. .. .. . (2s ) t2sν ν,ν (xν ) A2sν ,ν µ2sν ,ν Quadrature rules with multiple nodes 23 Remark 2. For the case γ = 0, if ∑2n ν=1 (2sν + 1)xν 6= `π, ` ∈ Z, then we choose tk,ν , k = 0, 1, . . . , 2sν , ν = 1, 2, . . . , 2n, in one of the following ways 2n x + ∑ (2si + 1)xi + 2sν xν tk,ν (x) = uk,ν (x) cos or i=1 i6=ν 2 2n x + ∑ (2si + 1)xi + 2sν xν tk,ν (x) = uk,ν (x) sin i=1 i6=ν 2 , which provides that tk,ν ∈ T∑2n ν=1 or tk,ν ∈ T∑2n ν=1 2n sν +n span cos ∑ sν + n x ν=1 2n sν +n span sin ∑ sν + n x , ν=1 k = 0, 1, . . . , 2sν , ν = 1, 2, . . . , 2n, respectively. 4.3 Numerical example In this section we present one numerical example as illustration of the obtained theoretical results. Numerical results are obtained using our proposed progressive approach. For all computations we use M ATHEMATICA and the software package OrthogonalPolynomials explained in [8] in double precision arithmetic. Example 1. We construct quadrature rule (11) for γ = 0, n = 3, σ = (3, 3, 3, 4, 4, 4), with respect to the weight function w(x) = 1 + cos 2x, x ∈ [−π, π). For calculation of nodes we use proposed progressive approach with only two steps, i.e., two iterative processes: for a = 1/2 (with 9 iterations) and for a = 1 (with 8 iterations). The nodes xν , ν = 1, 2, . . . , 6, are the following: −3.141592653589793, −2.264556388673865, −1.179320242581565, −0.1955027724705077, 0.8612188670819011, 2.178685249095223. The corresponding weight coefficients A j,ν , j = 0, 1, . . . , 2sν , ν = 1, 2, . . . , 6, are given in Table 1 (numbers in parentheses denote decimal exponents). 24 Gradimir V. Milovanović and Marija P. Stanić j A j,1 A j,2 0 1.676594372192496 7.305332409592605(−1) 1 −3.606295932640399(−3) −8.705060748567067(−2) 2 4.450266475268382(−2) 1.929684165460602(−2) 3 −6.813768854008439(−2) −9.825070962211830(−4) 4 2.575834718221877(−4) 1.050037121874394(−4) 5 −2.025745530496365(−7) −2.102329131299247(−6) 6 3.780621226622374(−7) 1.413143818423462(−7) j A j,4 A j,5 0 1.880646586862865 9.170128528217915(−1) 1 6.328640684710125(−2) −1.557489854117211(−1) 2 7.316493158593520(−2) 3.683058320865013(−2) 3 1.252054136886738(−3) −2.974661584717919(−3) 4 7.029326017474412(−4) 3.381006758789474(−4) 5 6.219410600844099(−6) −1.428080829445919(−5) 6 2.280678345497009(−6) 1.008832884300956(−6) 7 8.397304312672816(−9) −1.865855244589502(−8) 8 2.276058254314503(−9) 8.981189180630658(−10) A j,3 3.487838013502532(−1) 6.802235520408829(−2) 1.120920249826093(−2) 8.081866652580276(−4) 6.594073161399015(−5) 1.808851807084914(−6) 9.154154335491590(−8) A j,6 7.296144529929201(−1) 1.399304206896031(−1) 2.916174720939193(−2) 2.588432069654582(−3) 2.606855435340879(−4) 1.209621855269665(−5) 7.540115736887533(−7) 1.546523556082379(−8) 6.501031023479647(−10) Table 1 Weight coefficients A j,ν , j = 0, 1, . . . , 2sν , ν = 1, 2, . . . , 2n, for w(x) = 1 + cos 2x, x ∈ [−π, π), n = 3 and σ = (3, 3, 3, 4, 4, 4). 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